gate information is to determine how a signal is propagated in a special 2). Think of this signal as being a bit of information.

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1 one family of such curves that carry signals and provide a distinguished these roles. Even first order PDEs, which are actually wave-like have perbolic problems there is always a set of characteristic curves that play influence and domain of dependence. c2u, = 0, x > 0, < t < oo; u(0, t) = s(t), < t < 3. Solve the outgoing signal problem tions F and Gin the general solution (2.11) using the initial conditions (2.13). impulse. Contrast the solution with the case when f 0 and g = 0. Region of the solution surface and discuss the effect of giving a string at rest an initial Figure 2.3. displacement isf(x) = 0, and the initial velocity isg(x) = 1/( x 2). Plot 2. Calculate the exact solution to the Cauchy problem when c = 2, the initial 1. Derive d Alembert s formula (2.14) by determining the two arbitrary func So the signal breaks into two pieces, and they propagate in opposite dire c So the signal stays at the same place, but it spreads out and decreases in ill + 4kt u(x, t) = e2 (1+41t) 1 via u(x, t) = u(x, t) = O5(e_(_Ct) 2 + ex4ct) 2). Think of this signal as being a bit of information. case. For example, suppose the initial signal is a Gaussian function or bellshaped curve exp(x gate information is to determine how a signal is propagated in a special is retained. Parabolic, or diffusion, equations propagate signals at infinite herency in the wave form as it propagates, and therefore information parabolic and hyperbolic problems. Hyperbolic, or wave-like, equations The convection equation Ut + CU = 0, which is a wave-like equation, amplitude. Any information in the signal is eventually lost. :ion at (xo, to). This rm u 0. In hy Finall we point out again the important differences between speed; because the signals diffuse or smear out, there is a gradual loss of tions at speed c. The diffusion equation Ut = ku propagates the signal Section 1.2). coordinate sytem where the problem simplifies (recall the examples in propagate signals at a finite speed along characteristics; there is co information. A good way to understand how different equations propa propagates this signal via That is, it moves it at speed c without distortion. The wave equation =c2u moves it via 2). )lution outside this the wave equation fine a special coor ward in time to the regarding the charsignals forward in )und by tracing the y the initial values one. Looking at the itive characteristics, then the region of constant are paths affect the solution time along which ence of the interval a and x ct always zero outside :hat interval, so the n surface. :h velocity c. These z.z. The Jnbounded Domains 57

2 ct)/p Partial Differcntial Equations on Unbounded Domains 4. where s(t) is a known signal. Hint: Look for a right-traveling wave solution. The three-dimensional wave equation is 5. Solve the Cauchy problem c2au = 0, where is = u(x, z, t) and A is the Laplacian operator. For waves with spher ical symmetr is = u(p,, where p =.Jx2 + y2 + z2. In this special case the Laplacian is given by (Section 1.8) Au = u + u. By introducing a change of dependent variable U = pu, show that the general solution for the spherically symmetric wave equation is 1 u= (F(pct)+G(p+cfl). p Why do you think an outward-moving wave is = F(p amplitude? u1c 2u=0, xcr, t>0, 2 =c2(u + u) decays in -i u(x, 0) =e1, Ut(x, 0) = cosx, x R. Use a computer algebra program to graph the wave profile at t = 1, 2, 3. Take c= 1 6. In Section 1.7 we showed that any solution to Laplace s equation has the property that its value at a point is approximately the average of four nearby values surrounding the point. Can we make a statement about solutions to the wave equation? Consider any characteristic parallelogram (see Figure 2.4) whose sides are positive and negative characteristics, and let A, B, C, D be the vertices as shown. Show that any solution to the wave equation satisfies the relation u(a) + u(c) = u(b) + u(d). Figure 2.4. Characteristic parallelogram.

3 boo. Therefore, for each t > 0, w(x, t) = f(() lfr(y))g(x - t)dy. x = 0 the solution is 71 tisfies the Cauchy (x, t) are close. Let her. We would like i for Laplace s equa I close in the sense b this end, consider easily observe that ther problems such )n the initial and/or (ii) the solution is ions for various PDE ming. We say that a an initial boundary ) solve the problem, eed only speci1y the m (2J5)(2.16) does ty of continuous de e get the boundary in the boundary data lution should depend ehavior is disturbing, er all, we want to be ation (2.15)(2.16) is lation models steady :he data on the bound on Unbounded Domains this mean with regard to stability? showthatl u (x,t)u 2(x,t) I 8 +8Tforallx c R, 0 < t < T.Whatdoes since f G(x f6g(x t)dy = 6, Iu(x, t) v(x, t)i f (y) fr(y) II G(x 1. Show that the Cauchy problem for the backward diffusion equation, 2. Let u = u(x, y). Is the problem uxy = 0, 0 < x,y < 1, 3. Consider two Cauchy problems for the wave equation with different initial u (x, 0) = f (x), u(x, 0) = g (x), x I f (x) f 2(x) &, I g (x) g 2(x) IS 2, indices and not exponents). If for all x E R we have for i = 1, 2, where f,f2, g, and g2 are given functions (the superscripts are = x E R, 0 < t < T, data: square, a well-posed problem? Discuss. on the unit square, where the value of u is prescribed on the boundary of the for large n. u(x, t) = 1 + _en2tsinnx is unstable by considering the solutions u(x, 0) = ft), x R, u+u=0, xert>o, cj Lc closeness of the initial data implies closeness of the solution. t)dy = 1. Therefore, in the sense interpreted above, t) I The solution formula (2.8) gives

4 led Domains 75 of an odd solution to the positive real axis is the solution to the given ini tial boundary value problem. If this intuitive reasoning leaves the reader perplexed, then one can always verify analytically that the solutions we acetime have obtained by this reflection method are, in fact, solutions to the given where the problems. 2.24) is If the boundary condition (2.18) along x = 0 in the heat flow problem ion x > Ct (2.17)(219) is replaced by a Neumann condition y the g and can U(O, t) = 0, t > 0, embert s then the problem can be solved by extending the initial data to an even function. The same is true for the wave equation. We leave these calculations as exercises. (2.27) 0, (2.28) 1. Solve the problem < 0. = ku, x > 0, t > 0, u(0,t)=0, t >0, u(x, 0) = (x), x > 0, with an insulated boundary condition by extending an even function. The solution is to all of the real axis as u(x, y) = f[g(x t) + G(x + ](y)dy. s)ds 0 2. Find a formula for the solution to the problem u=ku, x>0,t>0, u(0,t)=0, t >0, eplaced by may write u(x, 0) = 1, x > 0. (s)ds, Sketch a graph of several solution profiles with k = Find the solution to the problem (2.29) 2 Utt = C U, x > 0, t > 0,.24) is given u(0, t) = 0, t > 0, )rx<ct. u(x 0) = xe X Ut(X 0) 0 x > 0. rn and wave hr these two Pick c = 0.5 and sketch several time snapshots of the solution surface to e restriction observe the reflection of the wave from the boundary.

5 ku = r) G(X r))f( Partial Differential Equations on Unbounded Domains 2.6. i Therefore, transformation of the dependent variable has changed the 2.6 problem into one with a homogeneous boundary condition, but a price was paidan inhomogeneit or source term g (t), was introduced into Laplac the PDE. In general, we can always homogenize the boundary condi- equati tions in a linear problem, but the result is an inhomogeneous PDE; SO ordina inhomogeneous boundary conditions can be traded for inhomogeneous an alg PDEs. operat We can solve (2.49)(2.51) for v(x, t) by formulating a Duhamel s prin- Let ciple. However, in Section 2.5 we will observe that Laplace transform grow t methods can also be applied to find the solution. means the La; The La 1. Write a formula given f for the solution to the problem U(s), ir sin x, x E R, t > o, variabi u(x, 0) where Ut(X, 0) = 0, X E R. is calle Graph the solution surface when c = 1. Laplac 2. Write a formula for many the solution s to the problem shortt SrnX, XE R, t > 0, system u(x,0)=0, XER. 3. Using Duhamel s principle, find a formula for the solution to the initial value returns problem for the convection equation Ut + cu = f(x, t), X E R, t > 0; u(x, 0) = 0, X E R. Hint: Look at the problem w(x,t;r)+cw(x,t;r)=o, XER,t >0; w(x,0;r)=flx,j, xer. Solve the problem Ut + 2u = xet, X E R, t > 0; U(x, 0) = 0, x E R. which i use which r 4. Formulate Duhamels principle and solve the initial boundary value problem The that it c Ut = ku +f(x, t), X > 0, t > 0, transfoi U(X,0)=0, X >0, u(0,t)0, t >0. The solution is u(x, t) = f f (G(X t + y t r)dydr. Formuh parts, ai equatlo;